Heat and Specific Heat

Q: How can I study Heat and Specific Heat effectively?

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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Q: What course covers Heat and Specific Heat?

Heat and Specific Heat is part of the AP Physics 2 course on Study Mondo, specifically in the Thermodynamics section. You can explore the full course for more related topics and practice resources.

Q: Are there practice problems for Heat and Specific Heat?

Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Heat transfer, specific heat capacity, and calorimetry

Written and reviewed by the Study Mondo Education TeamLast updated February 14, 2026

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🔥 Heat and Specific Heat

Heat vs. Temperature

Temperature: Measure of average kinetic energy of particles Heat (Q): Energy transferred between objects due to temperature difference

💡 Key Distinction: Temperature is a property; heat is energy in transit.

Heat flows spontaneously from hot to cold until thermal equilibrium is reached.

Units of Heat

Joule (J): SI unit of energy
Calorie (cal): Energy to raise 1 g of water by 1°C
- 1 cal = 4.186 J
Kilocalorie (kcal or Cal): Food "Calorie"
- 1 kcal = 1000 cal = 4186 J

Specific Heat Capacity

Specific heat (c) is the energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).

$Q = mc\Delta T$

where:

$Q$ = heat transferred (J)
$m$ = mass (kg)
$c$ = specific heat (J/kg·°C or J/kg·K)
$\Delta T$ = temperature change (°C or K)

Common Specific Heats:

Substance	c (J/kg·°C)
Water	4186
Ice	2090
Steam	2010
Aluminum	900
Steel	470
Copper	387
Lead	128

Water has the highest specific heat of common substances! This is why:

Coastal areas have moderate climates
Water is used in cooling systems
Land heats/cools faster than oceans

Sign Convention

$Q > 0$ : Heat absorbed (temperature increases)
$Q < 0$ : Heat released (temperature decreases)

$\Delta T = T_{final} - T_{initial}$

If $\Delta T > 0$ : temperature increases → heat absorbed If $\Delta T < 0$ : temperature decreases → heat released

Calorimetry

Calorimetry measures heat transfer using conservation of energy.

Principle: In an isolated system, heat lost = heat gained

$Q_{lost} + Q_{gained} = 0$

$\sum Q_i = 0$

Typical Setup:

Hot object placed in cold water
System isolated (insulated calorimeter)
Final temperature measured
Energy conservation: $Q_{hot} + Q_{cold} = 0$

Heat Capacity

Heat capacity (C) is heat needed to raise an object's temperature by 1°C:

$C = mc$

$Q = C\Delta T$

Units: J/°C

Difference:

Specific heat (c): Property of material (per kg)
Heat capacity (C): Property of specific object (total)

Methods of Heat Transfer

1. Conduction

Heat transfer through direct contact
Molecular collisions
Requires material medium
Example: Metal spoon in hot soup

2. Convection

Heat transfer by fluid motion
Hot fluid rises, cold sinks
Requires fluid (liquid or gas)
Example: Boiling water, ocean currents

3. Radiation

Heat transfer by electromagnetic waves
No medium required (works in vacuum)
All objects emit thermal radiation
Example: Sun warming Earth, heat lamps

Problem-Solving Strategy

Identify all objects exchanging heat
Set up energy conservation: $\sum Q = 0$
Write Q for each object: $Q = mc\Delta T$
Determine signs:
- Heating: $Q > 0$ , $\Delta T > 0$
- Cooling: $Q < 0$ , $\Delta T < 0$
Solve for unknown (usually final temperature or specific heat)
Check reasonableness: Final T should be between initial temperatures

Common Mistakes

❌ Using wrong sign for heat (lost vs. gained) ❌ Forgetting to convert units (g → kg, cal → J) ❌ Using Celsius in place of Kelvin inappropriately (ΔT is same, but not absolute T) ❌ Neglecting heat lost to surroundings (real calorimeters aren't perfectly insulated) ❌ Mixing up specific heat (c) and heat capacity (C)

📚 Practice Problems

1Problem 1easy

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Mass: $m = 2.0$ kg
Initial temp: $T_i = 20°\text{C}$
Final temp: $T_f = 80°\text{C}$
Specific heat: $c = 4186$ J/kg·°C

Find: Heat required $Q$

Solution:

Step 1: Calculate temperature change. $\Delta T = T_f - T_i = 80 - 20 = 60°\text{C}$

Step 2: Apply heat formula. $Q = mc\Delta T$ $Q = (2.0)(4186)(60)$ $Q = 502,320 \text{ J} = 502 \text{ kJ}$

Answer: 502 kJ of heat is required

This is equivalent to about 120 food Calories (kcal).

2Problem 2easy

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Mass: $m = 2.0$ kg
Initial temp: $T_i = 20°\text{C}$
Final temp: $T_f = 80°\text{C}$
Specific heat: $c = 4186$ J/kg·°C

Find: Heat required $Q$

Solution:

Step 1: Calculate temperature change. $\Delta T = T_f - T_i = 80 - 20 = 60°\text{C}$

Step 2: Apply heat formula. $Q = mc\Delta T$ $Q = (2.0)(4186)(60)$ $Q = 502,320 \text{ J} = 502 \text{ kJ}$

Answer: 502 kJ of heat is required

This is equivalent to about 120 food Calories (kcal).

3Problem 3medium

❓ Question:

A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Aluminum: $m_{Al} = 0.50$ kg, $T_{Al,i} = 100°\text{C}$ , $c_{Al} = 900$ J/kg·°C
Water: $m_w = 2.0$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C

Find: Final temperature $T_f$

Solution:

Step 1: Set up energy conservation. $Q_{Al} + Q_w = 0$ $m_{Al}c_{Al}\Delta T_{Al} + m_w c_w \Delta T_w = 0$

Step 2: Express temperature changes. $m_{Al}c_{Al}(T_f - T_{Al,i}) + m_w c_w(T_f - T_{w,i}) = 0$

Step 3: Substitute values. $(0.50)(900)(T_f - 100) + (2.0)(4186)(T_f - 20) = 0$ $450(T_f - 100) + 8372(T_f - 20) = 0$ $450T_f - 45,000 + 8372T_f - 167,440 = 0$ $8822T_f = 212,440$ $T_f = 24.1°\text{C}$

Verification:

Aluminum cools: $\Delta T_{Al} = 24.1 - 100 = -75.9°\text{C}$ ✓
Water warms: $\Delta T_w = 24.1 - 20 = 4.1°\text{C}$ ✓
Final T between initial temperatures ✓

Answer: Final temperature is 24.1°C

The water barely warms because it has much larger mass and specific heat!

4Problem 4medium

❓ Question:

A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Aluminum: $m_{Al} = 0.50$ kg, $T_{Al,i} = 100°\text{C}$ , $c_{Al} = 900$ J/kg·°C
Water: $m_w = 2.0$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C

Find: Final temperature $T_f$

Solution:

Step 1: Set up energy conservation. $Q_{Al} + Q_w = 0$ $m_{Al}c_{Al}\Delta T_{Al} + m_w c_w \Delta T_w = 0$

Step 2: Express temperature changes. $m_{Al}c_{Al}(T_f - T_{Al,i}) + m_w c_w(T_f - T_{w,i}) = 0$

Verification:

Aluminum cools: $\Delta T_{Al} = 24.1 - 100 = -75.9°\text{C}$ ✓
Water warms: $\Delta T_w = 24.1 - 20 = 4.1°\text{C}$ ✓
Final T between initial temperatures ✓

Answer: Final temperature is 24.1°C

The water barely warms because it has much larger mass and specific heat!

5Problem 5medium

❓ Question:

A 200 g aluminum cup (c = 900 J/(kg·°C)) contains 500 g of water (c = 4186 J/(kg·°C)) at 20°C. A 100 g piece of copper (c = 387 J/(kg·°C)) at 80°C is placed in the water. Find the final equilibrium temperature.

💡 Show Solution

Solution:

Given:

Al cup: m₁ = 0.200 kg, c₁ = 900 J/(kg·°C), T₁ = 20°C
Water: m₂ = 0.500 kg, c₂ = 4186 J/(kg·°C), T₂ = 20°C
Cu: m₃ = 0.100 kg, c₃ = 387 J/(kg·°C), T₃ = 80°C

Energy conservation: Heat lost by copper = Heat gained by water + aluminum

m₃c₃(T₃ - T_f) = m₁c₁(T_f - T₁) + m₂c₂(T_f - T₂)

Since T₁ = T₂ = 20°C: (0.100)(387)(80 - T_f) = (0.200)(900)(T_f - 20) + (0.500)(4186)(T_f - 20)

3096 - 38.7T_f = 180T_f - 3600 + 2093T_f - 41,860 3096 - 38.7T_f = 2273T_f - 45,460 48,556 = 2311.7T_f T_f = 21.0°C

The large mass and high specific heat of water dominates.

6Problem 6hard

❓ Question:

A 0.20 kg piece of unknown metal at 150°C is placed in 0.50 kg of water at 20°C. The final temperature is 25°C. What is the specific heat of the metal? Assume no heat is lost to surroundings. (c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Metal: $m_m = 0.20$ kg, $T_{m,i} = 150°\text{C}$ , $c_m = ?$
Water: $m_w = 0.50$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C
Final: $T_f = 25°\text{C}$

Find: Specific heat of metal $c_m$

Solution:

Step 1: Set up energy conservation. $Q_m + Q_w = 0$ $m_m c_m \Delta T_m + m_w c_w \Delta T_w = 0$

Step 2: Calculate temperature changes. $\Delta T_m = 25 - 150 = -125°\text{C}$ $\Delta T_w = 25 - 20 = 5°\text{C}$

Step 3: Solve for $c_m$ . $m_m c_m \Delta T_m = -m_w c_w \Delta T_w$ $c_m = -\frac{m_w c_w \Delta T_w}{m_m \Delta T_m}$ $c_m = -\frac{(0.50)(4186)(5)}{(0.20)(-125)}$ $c_m = -\frac{10,465}{-25}$ $c_m = 419 \text{ J/kg·°C}$

Comparison with known metals:

Copper: 387 J/kg·°C
Steel: 470 J/kg·°C

Answer: Specific heat is 419 J/kg·°C

This is close to copper (387) or possibly a copper alloy.

7Problem 7medium

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? Use c_water = 4186 J/(kg·°C).

💡 Show Solution

Solution:

Given: m = 2.0 kg, c = 4186 J/(kg·°C), ΔT = 80 - 20 = 60°C

Heat required: Q = mcΔT Q = (2.0)(4186)(60) Q = 5.02 × 10⁵ J or 502 kJ

8Problem 8hard

❓ Question:

💡 Show Solution

Given:

Metal: $m_m = 0.20$ kg, $T_{m,i} = 150°\text{C}$ , $c_m = ?$
Water: $m_w = 0.50$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C
Final: $T_f = 25°\text{C}$

Find: Specific heat of metal $c_m$

Solution:

Step 1: Set up energy conservation. $Q_m + Q_w = 0$ $m_m c_m \Delta T_m + m_w c_w \Delta T_w = 0$

Step 2: Calculate temperature changes. $\Delta T_m = 25 - 150 = -125°\text{C}$ $\Delta T_w = 25 - 20 = 5°\text{C}$

Comparison with known metals:

Copper: 387 J/kg·°C
Steel: 470 J/kg·°C

Answer: Specific heat is 419 J/kg·°C

This is close to copper (387) or possibly a copper alloy.

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❓ Frequently Asked Questions

What is Heat and Specific Heat?▾

Heat transfer, specific heat capacity, and calorimetry

How can I study Heat and Specific Heat effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Heat and Specific Heat study guide free?▾

Yes — all study notes, flashcards, and practice problems for Heat and Specific Heat on Study Mondo are free to access. No account is needed.

What course covers Heat and Specific Heat?▾

Heat and Specific Heat is part of the AP Physics 2 course on Study Mondo, specifically in the Thermodynamics section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Heat and Specific Heat?▾

Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Heat and Specific Heat

Try the Interactive Version!

🔥 Heat and Specific Heat

Heat vs. Temperature

Units of Heat

Specific Heat Capacity

Common Specific Heats:

Sign Convention

Calorimetry

Typical Setup:

Heat Capacity

Methods of Heat Transfer

1. Conduction

2. Convection

3. Radiation

Problem-Solving Strategy

Common Mistakes

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4medium

❓ Question:

5Problem 5medium

❓ Question:

6Problem 6hard

❓ Question:

7Problem 7medium

❓ Question:

8Problem 8hard

❓ Question:

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Thermodynamics

❓ Frequently Asked Questions